this should work just fine if you just need to show what time left, since JavaScript uses frames for its time you'll have get your End Time - The Time RN after that we can divide it by 1000 since apparently 1000 frames = 1 seconds, after that you can use the basic math of time, but there's still a problem to this code, since the calculation is static, it can't compensate for the different day total in a year (360/365/366), the bunch of IF after the calculation is to make it null if the time is lower than 0, hope this helps even though it's not exactly what you're asking :)

var now = new Date();
var end = new Date("End Time");
var total = (end - now) ;
var totalD =  Math.abs(Math.floor(total/1000));

var years = Math.floor(totalD / (365*60*60*24));
var months = Math.floor((totalD - years*365*60*60*24) / (30*60*60*24));
var days = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24)/ (60*60*24));
var hours = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24)/ (60*60));
var minutes = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60)/ (60));
var seconds = Math.floor(totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60 - minutes*60);

var Y = years < 1 ? "" : years + " Years ";
var M = months < 1 ? "" : months + " Months ";
var D = days < 1 ? "" : days + " Days ";
var H = hours < 1 ? "" : hours + " Hours ";
var I = minutes < 1 ? "" : minutes + " Minutes ";
var S = seconds < 1 ? "" : seconds + " Seconds ";
var A = years == 0 && months == 0 && days == 0 && hours == 0 && minutes == 0 && seconds == 0 ? "Sending" : " Remaining";

document.getElementById('txt').innerHTML = Y + M + D + H + I + S + A;

function DateDiff(b, e)
{
    let
        endYear = e.getFullYear(),
        endMonth = e.getMonth(),
        years = endYear - b.getFullYear(),
        months = endMonth - b.getMonth(),
        days = e.getDate() - b.getDate();
    if (months < 0)
    {
        years--;
        months += 12;
    }
    if (days < 0)
    {
        months--;
        days += new Date(endYear, endMonth, 0).getDate();
    }
    return [years, months, days];
}

[years, months, days] = DateDiff(
    new Date("October 21, 1980"),
    new Date("July 11, 2017")); // 36 8 20

var d1=new Date(2011,0,1); // jan,1 2011
var d2=new Date(); // now

var diff=d2-d1,sign=diff<0?-1:1,milliseconds,seconds,minutes,hours,days;
diff/=sign; // or diff=Math.abs(diff);
diff=(diff-(milliseconds=diff%1000))/1000;
diff=(diff-(seconds=diff%60))/60;
diff=(diff-(minutes=diff%60))/60;
days=(diff-(hours=diff%24))/24;

console.info(sign===1?"Elapsed: ":"Remains: ",
             days+" days, ",
             hours+" hours, ",
             minutes+" minutes, ",
             seconds+" seconds, ",
             milliseconds+" milliseconds.");

这段代码将返回两个日期的差值(以天为单位):

const previous_date = new Date("2019-12-23");
const current_date = new Date();

const current_year = current_date.getFullYear();
const previous_date_year = 
previous_date.getFullYear();

const difference_in_years = current_year - 
previous_date_year;

let months = current_date.getMonth();
months = months + 1; // for making the indexing 
// of months from 1

for(let i = 0; i < difference_in_years; i++){
months = months + 12;
}

let days = current_date.getDate();

days = days + (months * 30.417);

console.log(`The days between ${current_date} and 
${previous_date} are : ${days} (approximately)`);

像“差几天”这样的表达从来不像看起来那么简单。如果你有以下日期:

d1: 2011-10-15 23:59:00
d1: 2011-10-16 00:01:00

时间差2分钟，“天数差”应该是1还是0?类似的问题也出现在任何以月、年或其他形式表示的差异中，因为年、月和日的长度和时间不同(例如，夏令时开始的那一天比平时短1小时，比夏令时结束的那一天短2小时)。

这里是一个忽略时间的天数差函数，即对于上述日期，它返回1。

/*
   Get the number of days between two dates - not inclusive.

   "between" does not include the start date, so days
   between Thursday and Friday is one, Thursday to Saturday
   is two, and so on. Between Friday and the following Friday is 7.

   e.g. getDaysBetweenDates( 22-Jul-2011, 29-jul-2011) => 7.

   If want inclusive dates (e.g. leave from 1/1/2011 to 30/1/2011),
   use date prior to start date (i.e. 31/12/2010 to 30/1/2011).

   Only calculates whole days.

   Assumes d0 <= d1
*/
function getDaysBetweenDates(d0, d1) {

  var msPerDay = 8.64e7;

  // Copy dates so don't mess them up
  var x0 = new Date(d0);
  var x1 = new Date(d1);

  // Set to noon - avoid DST errors
  x0.setHours(12,0,0);
  x1.setHours(12,0,0);

  // Round to remove daylight saving errors
  return Math.round( (x1 - x0) / msPerDay );
}

这可以更简洁:

/* Return number of days between d0 and d1. ** Returns positive if d0 < d1, otherwise negative. ** ** e.g. between 2000-02-28 and 2001-02-28 there are 366 days ** between 2015-12-28 and 2015-12-29 there is 1 day ** between 2015-12-28 23:59:59 and 2015-12-29 00:00:01 there is 1 day ** between 2015-12-28 00:00:01 and 2015-12-28 23:59:59 there are 0 days ** ** @param {Date} d0 - start date ** @param {Date} d1 - end date ** @returns {number} - whole number of days between d0 and d1 ** */ function daysDifference(d0, d1) { var diff = new Date(+d1).setHours(12) - new Date(+d0).setHours(12); return Math.round(diff/8.64e7); } // Simple formatter function formatDate(date){ return [date.getFullYear(),('0'+(date.getMonth()+1)).slice(-2),('0'+date.getDate()).slice(-2)].join('-'); } // Examples [[new Date(2000,1,28), new Date(2001,1,28)], // Leap year [new Date(2001,1,28), new Date(2002,1,28)], // Not leap year [new Date(2017,0,1), new Date(2017,1,1)] ].forEach(function(dates) { document.write('From ' + formatDate(dates[0]) + ' to ' + formatDate(dates[1]) + ' is ' + daysDifference(dates[0],dates[1]) + ' days<br>'); });

加上@paresh mayani的答案，像Facebook一样工作-显示了以秒/分钟/小时/周/月/年为单位流逝了多少时间

var DateDiff = {

  inSeconds: function(d1, d2) {
        var t2 = d2.getTime();
        var t1 = d1.getTime();

        return parseInt((t2-t1)/1000);
    },


  inMinutes: function(d1, d2) {
        var t2 = d2.getTime();
        var t1 = d1.getTime();

        return parseInt((t2-t1)/60000);
    },

  inHours: function(d1, d2) {
        var t2 = d2.getTime();
        var t1 = d1.getTime();

        return parseInt((t2-t1)/3600000);
    },

    inDays: function(d1, d2) {
        var t2 = d2.getTime();
        var t1 = d1.getTime();

        return parseInt((t2-t1)/(24*3600*1000));
    },

    inWeeks: function(d1, d2) {
        var t2 = d2.getTime();
        var t1 = d1.getTime();

        return parseInt((t2-t1)/(24*3600*1000*7));
    },

    inMonths: function(d1, d2) {
        var d1Y = d1.getFullYear();
        var d2Y = d2.getFullYear();
        var d1M = d1.getMonth();
        var d2M = d2.getMonth();

        return (d2M+12*d2Y)-(d1M+12*d1Y);
    },

    inYears: function(d1, d2) {
        return d2.getFullYear()-d1.getFullYear();
    }
}







    var dString = "May, 20, 1984"; //will also get (Y-m-d H:i:s)
    
    var d1 = new Date(dString);
    var d2 = new Date();
    
    var timeLaps = DateDiff.inSeconds(d1, d2);
    var dateOutput = "";
    
    
    if (timeLaps<60)
    {
      dateOutput = timeLaps+" seconds";
    }
    else  
    {
      timeLaps = DateDiff.inMinutes(d1, d2);
      if (timeLaps<60)
      {
        dateOutput = timeLaps+" minutes";
      }
      else
      {
        timeLaps = DateDiff.inHours(d1, d2);
        if (timeLaps<24)
        {
          dateOutput = timeLaps+" hours";
        }
        else
        {
            timeLaps = DateDiff.inDays(d1, d2);
            if (timeLaps<7)
            {
              dateOutput = timeLaps+" days";
            }
            else
            {
                timeLaps = DateDiff.inWeeks(d1, d2);
                if (timeLaps<4)
                {
                  dateOutput = timeLaps+" weeks";
                }
                else
                {
                    timeLaps = DateDiff.inMonths(d1, d2);
                    if (timeLaps<12)
                    {
                      dateOutput = timeLaps+" months";
                    }
                    else
                    {
                      timeLaps = DateDiff.inYears(d1, d2);
                      dateOutput = timeLaps+" years";
                    }
                }
            }
        }
      }
    }
    
    alert (dateOutput);